Problem: Simplify the following expression and state the condition under which the simplification is valid. $q = \dfrac{-7y^2 + 105y - 378}{7y^3 - 35y^2 - 42y}$
Answer: First factor out the greatest common factors in the numerator and in the denominator. $ q = \dfrac {-7(y^2 - 15y + 54)} {7y(y^2 - 5y - 6)} $ $ q = -\dfrac{7}{7y} \cdot \dfrac{y^2 - 15y + 54}{y^2 - 5y - 6} $ Simplify: $ q = - \dfrac{1}{y} \cdot \dfrac{y^2 - 15y + 54}{y^2 - 5y - 6}$ Next factor the numerator and denominator. $ q = - \dfrac{1}{y} \cdot \dfrac{(y - 6)(y - 9)}{(y - 6)(y + 1)}$ Assuming $y \neq 6$ , we can cancel the $y - 6$ $ q = - \dfrac{1}{y} \cdot \dfrac{y - 9}{y + 1}$ Therefore: $ q = \dfrac{ -y + 9 }{ y(y + 1)}$, $y \neq 6$